Алгебра 8 класс учебник Макарычев, Миндюк ответы – номер 541

а) 25 = 26x − x² −26x + x² + 25 = 0 D = b² − 4ac = (−26)² − 4 ⋅ 1 ⋅ 25 = 676 − 100 = 576 > 0, уравнение имеет 2 корня
x₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{26+\sqrt{576}}{2 ⋅ 1} $$ = 26 + 24/2 = 50/2 = 25
x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{26-\sqrt{576}}{2 ⋅ 1} $$ = 26 − 24/2 = 2/2 = 1
x₁ = 25, x₂ = 1
б) 3t² = 10 − 29t 3t² − 10 + 29t = 0 D = b² − 4ac = 29² − 4 ⋅ 3 ⋅ (−10) = 841 + 120 = 961 > 0, уравнение имеет 2 корня
t₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-29+\sqrt{961}}{2 ⋅ 3} $$ = −29 + 31/6 = 2/6 = 1/3
t₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{-29-\sqrt{961}}{2 ⋅ 3} $$ = −29 − 31/6 = −60/6 = −10
t₁ = 1/3, t₂ = −10
в) y² = 4y + 96 y² − 4y − 96 = 0 D = b² − 4ac = (−4)² − 4 ⋅ 1 ⋅ (−96) = 16 + 384 = 400 > 0, уравнение имеет 2 корня
y₁ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{4+\sqrt{400}}{2 ⋅ 1} $$ = 4 + 20/2 = 24/2 = 12
y₂ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{4-\sqrt{400}}{2 ⋅ 1} $$ = 4 − 20/2 = −16/2 = −8
y₁ = 12, y₂ = −8
г) 3p² + 3 = 10p 3p² + 3 − 10p = 0 D = b² − 4ac = (−10)² - 4 ⋅ 3 ⋅ 3 = 100 − 36 = 64 > 0, уравнение имеет 2 корня
p₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{10+\sqrt{64}}{2 ⋅ 3} $$ = 10 + 8/6 = 18/6 = 3
p₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{10-\sqrt{64}}{2 ⋅ 3} $$ = 10 − 8/6 = 2/6 = 1/3
p₁ = 3, p₂ = 1/3
д) x² − 20x = 20x + 100 x² − 20x − 20x − 100 = 0 x² − 40x − 100 = 0 D = b² − 4ac = (−40)² − 4 ⋅ 1 ⋅ (−100) = 1600 + 400 = 2000 > 0, уравнение имеет 2 корня
x₁ = $$ =\frac{-b+\sqrt{D}}{2 a}=\frac{40+\sqrt{2000}}{2 ⋅ 1}=\frac{40+20 \sqrt{5}}{2}= $$ = 20 + 10$$\sqrt{5}$$
x₂ = $$ \frac{-b+\sqrt{D}}{2 a}=\frac{40-\sqrt{2000}}{2 ⋅ 1}=\frac{40-20 \sqrt{5}}{2} $$ = 20 − 10$$\sqrt{5}$$
x₁ = 20 + 10$$\sqrt{5}$$, x₂ = 20 − 10$$\sqrt{5}$$
е) 25x² − 13x = 10x² − 7 25x² − 13x − 10x² + 7 = 0 15x² − 13x + 7 = 0 D = b² − 4ac = (−13)² − 4 ⋅ 15 ⋅ 7 = 169 - 420 = -251 < 0, уравнение не имеет корней